Heiro: Good morning, Archimedes! I bet you 1000 drachma that you can’t answer a simple maths problem!

Archimedes: You’re on.

Heiro: Did you know that every sum of consecutive cubes is also a difference of squares?

Archimedes: Indeed, as everyone knows, each and every sum of consecutive cubes is also a difference of consecutive squares. Was that the question?

Heiro: Not so fast, Archimedes! What if the sum and difference are a prime number?

Archimedes (yawning): Unlikely, since any sum of cubes has at least two factors…

Heiro: Let’s make it a difference of consecutive cubes, then?

Archimedes: There is no reason why such a difference should not be prime, as often as occasion demands. For example, 2^3-1^3 = 7. Have I won yet?

Heiro: I'm just working up to it. What about a difference of consecutive cubes, which is not only prime, but is also a sum of consecutive squares?

Archimedes: Too easy! 61 = 5^3-4^3=5^2+6^2. Can I have my money now?

Heiro: Very well, then, here's my question – just to make it interesting, let’s also stipulate that a side of the larger cube must have at least 4 distinct prime factors – so it must be at least 2*3*5*7 = 210, say? Just how big would that cube have to be?